Union with simply connected set

Question

Let $U$ and $V$ be open and path-connected sets in $X$ such that $U$ is simply connected and $U \cap V \neq \emptyset$ is path-connected. Is it true that $\pi_1(U \cup V) = \pi_1(V)$? If not, is there a condition under which this is true? (I believe this would follow from Van-Kampen, but not sure)

Answer 1

The theorem of Seifert - van Kampen says that $$\pi_1(U \cup V) \approx (\pi_1(U) * \pi_1(V))/N$$ where $*$ denotes the free product and $N$ is the normal subgroup generated by all elements of the form $$(i_U)_*(g)^{-1}(i_V)_*(g)$$ where $g \in \pi_1(U \cap V)$ and $i_U : U \cap V \to U$ and $i_V : U \cap V \to V$ are the inclusion maps. Note that $(i_U)_*(g)^{-1}(i_V)_*(g)$ is a word with $(i_U)_*(g) \in \pi_1(U)$ and $(i_V)_*(g) \in \pi_1(V)$.

We have $\pi_1(U) = 0$, thus

• $\pi_1(U) * \pi_1(V) = \pi_1(V)$
• $N$ is the normal subgroup generated by the image of $(i_V)_*$.